Environmental Engineering Reference
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J
( q ) has its maximum, to a nearby commensurable value Q c . Hence
the hexagonal anisotropy couples the helical magnetic structure to the
lattice, and it may induce continuous or abrupt changes of the ordering
wave-vector as a function of temperature, as discussed, for instance, by
Bak (1982). In Ho, 12 Q 0 is close to 4 π/c , and the hexagonal anisotropy
is large at low temperatures. Experimental investigations have shown
that a number of commensurable values of Q are stabilized in this sys-
tem, as we shall discuss in more detail in the last section of this chapter.
2.1.4 Longitudinally ordered phases
When B 2 is negative, as in Er and Tm, χ ζζ ( Q ) is the component of the
susceptibility which diverges at the highest temperature, and the high-
temperature expansion predicts that 2 A ζ −J
( Q ) vanishes at a critical
temperature determined by
( Q ) 1
) B 2 /k B T N .
k B T N 3
5
2
)( J + 2
J ( J +1)
J
( J
(2 . 1 . 31)
Just below this temperature, only the component σ ζ at the wave-vector
Q is non-zero and, from the free energy expansion (2.1.24), ∂f/∂σ ζ =0
determines the relative magnetization as
σ ζ ( Q )= σ Q = J
1 / 2
2 A ζ
3 J 2 B ζζ
( Q )
.
(2 . 1 . 32)
The free energy is independent of the phase ϕ = ϕ ζ ,soweset ϕ =0. If
we add another Fourier component with q
=
± Q :
= Q cos ( Q · R i )+ q cos ( q · R i + ϕ )
J
(2 . 1 . 33)
then, if m Q ±
n q is different from a reciprocal lattice vector, where m
and n are integers and m + n =
±
4, the free energy is
J 2 {
J 4 B ζζ 3 σ 4 Q +3 σ q
+12 σ 2 Q σ q +4 σ 3 Q σ q δ q 3 Q cos ϕ +4 σ Q σ q δ 3 q Q cos 3 ϕ .
(2 . 1 . 34)
σ q + 8
f = 4
σ 2 Q +
2 A ζ −J
( Q )
}
{
2 A ζ −J
( q )
}
1
This result shows that, if q =3 Q or q =
3 Q , there is an extra fourth-
order contribution to the free energy ( q →− q represents the same
structure with ϕ →−
ϕ ). Of these two special cases, the one where
q =3 Q is the most interesting, because the extra term is linear in σ 3 Q .
This means that the third harmonic appears simultaneously with the
basic Fourier component at Q .
Minimizing the free energy given by
(2.1.34), we find
J 2 B ζζ
J ( Q ) −J (3 Q ) σ 3 Q
ϕ = ϕ + π,
σ 3 Q =
(2 . 1 . 35 a )
;
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